Translate the congruent number problem into elliptic curve, we conclude that an integer $n\in\mathbb Z^+$ is area of a right triangle with $a,b,c\in\mathbb{Q}$ if and only if the corresponding elliptic curve has positive rank, viz. the curve $E: y^2=x^3-n^2x$ over $\mathbb{Q}.$
I'm fascinated, since I work in Gaussian rationals: can we also say $E/\mathbb{Q}$ is finite (infinite) if and only if $E/\mathbb{Q}(i)$ finite (infinite)? Maybe a conjugate of point argument? This interesting since not always true for elliptic curves, but is true for this one!
Yes, this is true because in this particular case we have $\text{rank}_\mathbb{Z} E(\mathbb{Q}(i)) =2\text{rank}_\mathbb{Z} E(\mathbb{Q}) $. The intuition for this is that for every point $(x_0,y_0)\in E(\mathbb{Q})$, there is also a point $(-x_0,iy_0)\in E(\mathbb{Q}(i))$.
In general, one has the following result.
Theorem. Let $E/\mathbb{Q}$ be an elliptic curve given by a Weierstrass equation of the form $y^2=f(x)$, and let $d$ be a square-free integer. Then, $$E(\mathbb{Q}(\sqrt{d})) \cong E(\mathbb{Q}) \oplus E^d(\mathbb{Q}),$$ where $E^d/\mathbb{Q}$ is the elliptic curve given by $dy^2=f(x)$.
In your case, $E_n: y^2=x^3-n^2x$ and $d=-1$, so $E_n^{-1} : -y^2=x^3-n^2x$ is isomorphic to $E$ over $\mathbb{Q}$, with a map $E_n^{-1}$ to $E_n$ given by $(x,y)\mapsto (-x,y)$. Hence, the theorem above implies that $$E_n(\mathbb{Q}(i)) \cong E_n(\mathbb{Q}) \oplus E_n^{-1}(\mathbb{Q})\cong E_n(\mathbb{Q}) \oplus E_n(\mathbb{Q}).$$ In particular, we have $\text{rank}_\mathbb{Z} E(\mathbb{Q}(i)) =2\text{rank}_\mathbb{Z} E(\mathbb{Q})$ as we had previously claimed.
The isomorphism $\psi: E_n(\mathbb{Q})\oplus E_n^{-1}(\mathbb{Q}) \to E_n(\mathbb{Q}(i))$ is given by
if $P\in E_n(\mathbb{Q})$, then $\psi(P)=P$,
if $Q=(x_0,y_0)\in E_n^{-1}(\mathbb{Q})$ then $$Q=(x_0,y_0) \mapsto (x,iy)\in E_n(\mathbb{Q}(i)),$$
and extend the map to $E_n(\mathbb{Q})\oplus E_n^{-1}(\mathbb{Q})$ by linearity, i.e., $\psi(P,(x_0,y_0)) = P + (x_0,iy_0)$.